Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}6x+3y &= 4 \\ 2x+6y &= -7\end{align*}$
Solution: Begin by moving the $x$ -term in the second equation to the right side of the equation. $6y = -2x-7$ Divide both sides by $6$ to isolate $y$ $y = {-\dfrac{1}{3}x - \dfrac{7}{6}}$ Substitute this expression for $y$ in the first equation. $6x+3({-\dfrac{1}{3}x - \dfrac{7}{6}}) = 4$ $6x - x - \dfrac{7}{2} = 4$ Simplify by combining terms, then solve for $x$ $5x - \dfrac{7}{2} = 4$ $5x = \dfrac{15}{2}$ $x = \dfrac{3}{2}$ Substitute $\dfrac{3}{2}$ for $x$ back into the top equation. $6( \dfrac{3}{2})+3y = 4$ $9+3y = 4$ $3y = -5$ $y = -\dfrac{5}{3}$ The solution is $\enspace x = \dfrac{3}{2}, \enspace y = -\dfrac{5}{3}$.